Solid solute
Hfusion >0 --------->
Solution
Ideal solutions. An ideal solution is defined as the one which can be formed from its components with no evolution or absorption of heat. Altough liquids exist for the existence of interaction between molecules, for many liquid diluted solutions the assumption above reasonably holds and the study of solution can be greatly simplified. From thermodynamics we know that the chemical equilibrium is the results of two conflicting forces: the tendency of a system to reach a state of minimum energy and the tendency of a system to reach a state of maximum entropy (chaos, disorder). Since we assume that under mixing no heat is evolved or absorbed the only force we should consider is entropy. When two liquids are mixed, the desorder (entropy) always increases. In fact, in the pure liquid a molecule is surrounded by other molecules of the same kind while in the solution we can not predict what is the identity of neighbors. Therefore, since the energetic state of pure liquids and that of the ideal solution is the same there is no resistance to the tendency of the system to reach a state of maximum chaos (disorder) and thus no equilibrium can exist. Terefore, two liquids forming an ideal solution can be mixed in any proportions.
Consider now the case of a solid solute, the dissolution process requires energy to breakdown attractive forces between solute molecules. The process of dissolution can be represented as follows:
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Solid solute |
Hfusion >0 ---------> |
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Solution |
The liquid solute is not in a state of minimum energy and this contrasts with the tendency of the system to reach a state of maximum disorder. Therefore, the solute can not mixed in any proportion with the solvent. For example, if we add slowely silver chloride to one liter of water we can observe the formation of an homogeneous solution (all solid dissolves) until 0.0024 grams are added. When this amount is reached, no further dissolution of AgCl occurs and solid salt accumulates in the liquid. This kind of solution is said saturated and the maximum amount of the salt which can be dissolved is called solubility of the salt. Therefore the solubility of AgCl is 0.0024 grams/liter.
Non-ideal solutions. In a non-ideal solution the interaction between solute and solvent can lead to evolution or absorption of heat. When heat is evolved the formation of the solution is both energetically and entropically favored and then the two liquids can be mixed in any proportions. On the contrary, when heat is absorbed the formation of the solution depends on the balance of the two conflicting forces (entropy and energy). This means that the two liquids can not be mixed in any proportions.
Analogously, in the case of solid solutes, we should expect that an evolution of heat favors the solubility while the absorption of heat will be unfavourable for solubility.
However, in practice, is difficult to qualitatively predict the solubility of a solute using the generalizations made above. In fact, for molecules strongly interacting, is difficult to predict the changes in energy and entropy.
If we add slowely silver chloride to water we can observe the formation of an homogeneous solution (all solid dissolves) until the concentration is less than 1.7 x 10-5 M. When this concentration is reached, no further dissolution of AgCl occurs and solid salt accumulates in the liquid. In other words, for further additions of the salt the concentration of the dissolved salt remain unchanged (1.7 x 10-5 M). This kind of solution is said saturated and the maximum amount of the salt which can be dissolved is called the concentration of saturation (solubility) of the salt.
Let consider the reaction of decomposition of calcium carbonate (solid) into calcium oxide (solid) and carbonic anhydride (gas):
| CaCO3(s) = CaO(s) + CO2(g) |  [CO2] [CaO] K' = ------------- [CaCO3] |
Since the concentration of a solid is constant we can include the concentration of CaCO3(s) and CaO(s) in the equilibrium constant thus obtaining
[CaCO3]
1. K'--------- = [CO2] = K
[CaO]
Therefore the equilibrium constant for the decomposition of calcium carbonate can expressed only in terms of CO2 concentration (or partial pressure) at equilibrium.
Consider now the dissolution of iodine in carbon tetrachloride:
| I2(s) = I2(l) | [I2]l K' = ------- [I2]s |
The concentration of solid I2 is constant and including it in the constant we have that
K = [I2] = solubility.
Therefore, the equilibrium constant for a reaction of dissolution is equal to the solubility of the solute in the solvent i.e., the maximum amount of solute which can be dissolved, at a given temperature, in the solvent.
Salts, with few exception (e.g., HgCl2), are strong electrolytes, i.e., full dissociated in solutions. However, salts greatly differ for their solubility in water. For example, KCl is a very water soluble salt while AgCl is a slight water soluble salt. For a slight water soluble salt, in practical problems of chemical analysis, is quite common to reach concentrations at which solid salt coexists with the solubilized salt, i.e. there is an equilibrium such the following:
1. AgCl(solid) Ag+(aqueous) + Cl-(aqueous) |
[Ag+] [Cl-] K = ------------ [AgCl(solid)] |
Including the solid concentration (wich is constant) in K, we get
2. K [AgCl(solid)] = [Ag+] [Cl-] = Ksp
The product of K with the solid concentration is a new constant generally indicated as Ksp. The equation is generally called the ion product or the solubility product.
It should be evident that if the product of the concentrations exceed the value of Ksp some salt must precipitate to restablish the equilibrium. On the other hand, no precipitation occur and no solid is present if the product of the concentrations is less than Ksp.
Example I. The value of Ksp for AgCl is 2.8 x 10-10, calculate the concentrations of Ag+, Cl- and the solubility of the salt.
From the stoichiometry of the reaction it is evident that
| [Ag+] = [Cl-] | Ksp = [Ag+] [Cl-] = [Ag+]2 = 2.8 x 10-10 |
from which [Ag+] = (2.8 x 10-10)1/2 = 1.7 x 10-5 M and of course [Cl-] = 1.7 x 10-5 M.
Since the concentration of Ag+ (or Cl-) in the solution must match the amount of AgCl dissociated then the solubility of of silver chloride in water is 1.7 x 10-5 M, i.e., only 1.7 x 10-5 moles/liter dissociates.
Example II. Calculate the solubility of calcium fluoride (CaF2, Ksp = 1.7 x 10-10).
From the stoichiometry of the reaction
| CaF2 Ca2+ + 2F- |
Ksp = [Ca2+][F-]2 =1.7 x 10-10 |
it results that
| F- = 2Ca2+ | Ksp = [Ca2+] [2Ca2+]2 = 4[Ca2+]3 = 1.7 x 10-10 |
from which [Ca2+] = 3.5 x 10-4 M. Therefore the solubility of CaF2 in water is 3.5 x 10-4 M.
Example III. Calculate the solubility of silver chloride in a 0.1 M AgNo3 solution. Consider that AgNo3 is a very soluble salt.
The concentration of Ag+ at the equilibrium should be the sum of Ag+ coming from silver chloride and that coming from silver nitrate, i.e.,
[Ag+]total = [Ag+]nitrate + [Ag+]chloride
Since silver nitrate is very soluble, (i.e., there is no equilibrium between solids and ions) then the concentration of [Ag+]nitrate is O.1 M. The solubility of silver chloride in pure water is 1.7 x 10-5 M, but according to LeChatelier' principle (common ion effect) it will be less for the presence [Ag+]nitrate. It follows that with a good approximation [Ag+]total = [Ag+]nitrate = 0.1 M. Now we can calculate the concentration of Cl- at equilibrium:
Ksp = [Ag+] [Cl-] = O.1[Cl-] = 2.8 x 10-10[Cl-] = 2.8 x 10-10/ O.1 = 2.8 x 10-9 M
Thus the amount of AgCl dissociated (the solubility) is 2.8 x 10-9 moles/liter.
Let's consider a solution containing 0.1 M Cl- and 0.01 M CrO4-- at which a solution of AgNO3 is slowely added. The solubility product of silver chloride and silver chromate are:
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From the value of Ksp of AgCl, it can be calculated that when Cl- is 0.1 M:
[Ag+] at equilibrium is 2.8 x 10-9 M
This means that silver chloride will begin to precipitate from the solution when we add enough silver nitrate to reach this concentration.
For the case of chromate we can calculate that [Ag+] at equilibrium is 1.4 x 10-5 M. Thus if we add slowely a solution of silver nitrate to a solution containing 0.1 M Cl- and 0.01 M CrO4-- , AgCl start to precipitate when [Ag+] is 2.8 x 10-9 M. Further addition of silver nitrate will cause more precipitation of AgCl but no precipitation of of Ag2CrO4 until the concentration of [Ag+] becomes 1.4 x 10-5 M. Thus, by this approach we could separate AgCl from Ag2CrO4.
An interesting problem is to calculate if the precipitation of AgCl is completed when Ag2CrO4 starts to precipitate. Calculations show that practically all AgCl is precipitated when chromate starts to precipitate (see problems).
A solution of chromate ions is bright yellow but a precipitate of silver chromate is dark red. This means, according to the discussion on the selective precipitation above, that when a dark red precipitate appears all the AgCl has been precipitated from the solution.
Consider now a solution with an unknown concentration of Cl- our task is to determine the concentration of Cl- by using a solution of AgNO3 with known concentration. First, we fill a uniform-bore glass tube with fine volume gradations and a stopcock at the bottom (burette) with the solution of silver nitrate. Second, we add a small amount of chromate ion to the solution of Cl-. Now we slowely add the solution of silver nitrate to the solution of chloride ion (which should be continuously agitated) and as soon as a dark red color is observed we stop the dispensing of silver nitrate. Finally, we read the volume of nitrate added from which we can calculate the moles of Ag+ added. Since the reaction considered is
Ag+ + Cl- = AgCl
the moles of Ag+ are equals to the moles of Cl- and the concentration of Cl- can be calculated.
This very powerfull analytical technique is called titration, AgNO3 is called the titrant and Ag2CrO4 which indicates the endpoint of the titration is known as indicator.